Multiple Choice
In the context of Raoult's Law, what is the term for the pressure exerted by vapor molecules on the sides of a container at equilibrium?
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14. Solutions
Vapor Pressure Lowering (Raoult's Law)
Multiple Choice
Given four solutions at the same temperature, which is expected to have the lowest vapor pressure according to Raoult's Law?
A
A solution of 1.0 mol NaCl in 1.0 kg water
B
Pure water
C
A solution of 0.5 mol glucose in 1.0 kg water
D
A solution of 0.5 mol NaCl in 1.0 kg water
Verified step by step guidance1
Recall Raoult's Law, which states that the vapor pressure of a solution, \(P_{\text{solution}}\), is related to the mole fraction of the solvent, \(X_{\text{solvent}}\), and the vapor pressure of the pure solvent, \(P^0_{\text{solvent}}\), by the equation:
\[P_{\text{solution}} = X_{\text{solvent}} \times P^0_{\text{solvent}}\]
Understand that adding a solute to a solvent lowers the mole fraction of the solvent, thus lowering the vapor pressure compared to the pure solvent. The more solute particles present, the lower the mole fraction of the solvent and the lower the vapor pressure.
Recognize that NaCl is an ionic compound that dissociates into ions in solution. Specifically, 1 mole of NaCl produces 2 moles of particles (Na\(^+\) and Cl\(^-\)), while glucose does not dissociate and remains as 1 mole of particles per mole of glucose.
Calculate the total moles of solute particles for each solution:
- For 1.0 mol NaCl, total particles = 1.0 mol \(\times\) 2 = 2.0 mol
- For 0.5 mol NaCl, total particles = 0.5 mol \(\times\) 2 = 1.0 mol
- For 0.5 mol glucose, total particles = 0.5 mol (no dissociation)
- Pure water has 0 mol solute particles.
Since the solution with the greatest number of solute particles lowers the mole fraction of water the most, it will have the lowest vapor pressure. Therefore, the solution of 1.0 mol NaCl in 1.0 kg water is expected to have the lowest vapor pressure.
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